"An archer releases an arrow from a shoulder height of 1.30 m. When the arrow hits the target 18m away, it hits point A. When the target is removed, the arrow lands 45m away. Find the maximum height of the arrow along its parabolic path." How do you find the maximum?

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perl · Answer

this question was asked yesterday

anonymous · Answer

$$y - y _{o}= vt + \frac{ 1 }{ 2 }at ^{2}$$
$$t = \sqrt{\frac{ 2(-y) }{ g }}$$
$$t = \sqrt{\frac{ 2(-1.3m) }{ -9.8m/s ^{2} }}$$= 0.52s

Now plug in the time  for the x direction.
$$x = vt$$
$$v = \frac{ x }{ t }$$
$$v  = \frac{ 45m }{ 0.52 s }= 86.5m/s$$
Use vx to find the angle
vx cos(x) 

Then find the y component of v.
vy sin(x)

Then plug in vy back into the first equation.

anonymous · Answer

Thank you all:)

perl · Answer

here http://openstudy.com/users/perl#/updates/54614094e4b0c2835b86c74b

perl · Answer

you don't need to use gravity in the problem,

perl · Answer

you are given three points and you want to fit a quadratic curve to it

perl · Answer

@Jhannybean  also the question is incomplete, he did not define 'point A'

anonymous · Answer

Jhannybean did not do this right.

Shes using gravity in the horizontal direction.
This is a projectile, the only accelerate is going on is gravity which is in the vertical direction.

jhannybean · Answer

|dw:1415726259961:dw|

anonymous · Answer

Looking back the question is unclear. It gave us two different ranges. The max height when the range is 18m away or the max height when the target is gone, and the range is 45m away?

jhannybean · Answer

typo. |dw:1415727045399:dw|